Well don't worry about what the value of \(a\) yet (although you are right in that it is negative and the gravitational acceleration constant, which is \(-g = -9.8 \text{m}/\text{s}^2\).
Solving for \(v(t)\) should be easy. If the derivative (acceleration) is constant, then \(v(t) = a t + C\) right?

Ok. well, we still have on more time-integral left. since this isn't for you, I'll just give you the solution: \(y = y_0 + v_0 t + a t^2/2\) I trust you can find approp. values for the constants, right?